Calculate the molar concentrations of all phosphate species in H3PO4 in a solution with
pH = 2.25, given that the concentration of all four forms of phosphate total 15 mM.
Convert pH to (H^+).
Work this by the fractional method.
alpha o = (H^+)^3/D
alpha 1 = (H^+)^2*k1/D
alpha 2 = (H^+)^k1k2/D
alpha 3 = k1k2k3/D
D is
(H^+)^3 + (H^+)^2*k1 + (H^+)*k1k2 + k1k2k3
Then (H3PO4) = alphao*15 mM
(H2PO4^-) = alpha1*15 mM
etc.
To calculate the molar concentrations of all phosphate species in H3PO4, we need to consider the dissociation reactions of phosphoric acid.
Phosphoric acid (H3PO4) can dissociate into four different species:
H3PO4 ⇌ H+ + H2PO4-
H2PO4- ⇌ H+ + HPO42-
HPO42- ⇌ H+ + PO43-
Given that the concentration of all four forms of phosphate total 15 mM and the pH of the solution is 2.25, we can use the following relationships to calculate the molar concentrations:
[H3PO4] + [H2PO4-] + [HPO42-] + [PO43-] = 15 mM
pH = -log[H+]
Since H3PO4 is a weak acid, we can approximate that most of it will remain undissociated at pH 2.25. Therefore, we can assume that [H3PO4] ≈ 15 mM.
Using the equilibrium constant expression for the first dissociation reaction, we can set up the relationship between [H+], [H3PO4], and [H2PO4-]:
Ka1 = ([H+][H2PO4-])/[H3PO4]
The value of Ka1 for phosphoric acid is approximately 7.5 x 10^-3.
At pH 2.25, [H+] can be calculated as follows:
[H+] = 10^(-pH)
= 10^(-2.25)
Now we can rearrange the equilibrium constant expression to solve for [H2PO4-]:
[H2PO4-] = (Ka1 * [H3PO4])/[H+]
= (7.5 x 10^-3 * 15 mM) / (10^(-2.25) M)
Similarly, we can use the equilibrium constant expression for the second dissociation reaction and the fact that [HPO42-] = [H+] to calculate [HPO42-]:
Ka2 = ([H+][HPO42-])/[H2PO4-]
[HPO42-] = (Ka2 * [H2PO4-])/[H+]
Finally, we can use the equilibrium constant expression for the third dissociation reaction and the fact that [PO43-] = [H+] to calculate [PO43-]:
Ka3 = ([H+][PO43-])/[HPO42-]
[PO43-] = (Ka3 * [HPO42-])/[H+]
By substituting the values of [H2PO4-] and [HPO42-] into the equation above, we can find [PO43-].
Please note that to obtain the exact values of the concentrations, you would need to plug in the numerical values for [H+], [H3PO4], and the dissociation constants Ka1, Ka2, and Ka3 into the equations.
To calculate the molar concentrations of all phosphate species in H3PO4, we need to understand the dissociation of phosphoric acid (H3PO4) in water.
H3PO4 is a triprotic acid, meaning it can donate three protons (H+ ions). It undergoes three ionization steps:
H3PO4 ⇌ H+ + H2PO4- (K1)
H2PO4- ⇌ H+ + HPO4-2 (K2)
HPO4-2 ⇌ H+ + PO4-3 (K3)
Given the pH of 2.25, we can assume that the majority of H3PO4 has dissociated, forming H+ ions and anionic phosphate species.
Let's denote the molar concentrations of H3PO4, H2PO4-, HPO4-2, and PO4-3 as [H3PO4], [H2PO4-], [HPO4-2], and [PO4-3] respectively.
Since the concentration of all four species combined is given as 15 mM, we can express it as:
[H3PO4] + [H2PO4-] + [HPO4-2] + [PO4-3] = 15 mM (Equation 1)
Now, let's consider the dissociation of each species.
For H3PO4:
[H3PO4] ⇌ H+ + H2PO4-
We can assume that the concentration of [H3PO4] is equal to the initial concentration since the other species are formed through its dissociation. Therefore, [H3PO4] = 15 mM.
For H2PO4-:
[H2PO4-] ⇌ H+ + HPO4-2
[H3PO4] can donate a proton to form H2PO4-. Thus, the concentration of [H2PO4-] is equal to the concentration of H3PO4 dissociated. Therefore, [H2PO4-] = [H+].
For HPO4-2:
[HPO4-2] ⇌ H+ + PO4-3
[H2PO4-] can donate a proton to form HPO4-2. Thus, the concentration of [HPO4-2] is equal to the concentration of H2PO4- dissociated. Therefore, [HPO4-2] = [H+] = [H2PO4-].
Now, let's rewrite Equation 1 using the derived expressions for each species:
[H3PO4] + [H2PO4-] + [HPO4-2] + [PO4-3] = 15 mM
15 mM + [H+] + [H+] + [H+] = 15 mM
3[H+] = 15 mM
[H+] = 5 mM
Since [H3PO4] = [H2PO4-] = [HPO4-2] = [H+], we can substitute [H+] with [H3PO4] = [H2PO4-] = [HPO4-2] = 5 mM.
Therefore, the molar concentrations of all phosphate species are as follows:
[H3PO4] = 5 mM
[H2PO4-] = 5 mM
[HPO4-2] = 5 mM
[PO4-3] = unknown (since it is not directly provided)
Note: The concentration of [PO4-3] cannot be determined solely based on the given pH and total concentration of the four species. Additional information or assumptions are needed to calculate [PO4-3].